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Math problem set Grade 10 Careers & CTE

Askwith Heights Pressure Zone: Distribution Operator Math

A grade 10 careers math set following a distribution operator through a low-pressure complaint and a flushing run, using head, pressure, flow, and pipe volume.

Objectives

  1. Students will convert elevation difference into static pressure and back again using standard operator constants.
  2. Students will find flow in a main from pipe area and velocity, then express it in three different units.
  3. Students will compute the volume of a length of main and the time needed to flush it at a stated flow.

Vocabulary

Pressure zone
A pressure zone is a section of a water system that a utility manages separately because it sits at a different elevation than the rest of the town. Utilities create pressure zones so that water pressure stays safe and steady for everyone, whether they live at the bottom of a hill or near the top.

Also: static head, elevated tank, velocity, flushing

Teacher overview

A distribution operator who takes a low-pressure complaint answers it with arithmetic before ever touching a valve, because the pressure at any address in a gravity zone is set by how far that address sits below the water surface in the tank. This set follows one operator through a shift at the fictional Merribourne Regional Water, first checking pressures in the Askwith Heights zone and then planning a flushing run on one main. The misconception it is built to correct is the habit of using the height of the tank itself, or the tank's ground elevation, instead of the difference between the water surface elevation and the elevation of the point being checked. Operators use the constants that appear on state certification formula sheets: one foot of head equals 0.433 psi, and one psi equals 2.31 feet of head.

Askwith Heights Pressure Zone: Distribution Operator Math — Grade 10 Careers & CTE
Name: ____________________________ Date: ______________

Problems

The fictional Merribourne Regional Water runs the Askwith Heights pressure zone from one elevated tank. The water surface in that tank sits at elevation 1,186.0 feet. The zone is fed by a 12-inch main, which is 1.0 foot in diameter. Use these constants: 1 foot of head equals 0.433 psi, 1 psi equals 2.31 feet of head, 1 cubic foot per second equals 448.8 gallons per minute, 1 million gallons per day equals 694 gallons per minute, and 1 cubic foot holds 7.48 gallons. Use 0.785 times diameter squared for the area of a round pipe. Round each answer to the place shown in the question.

  1. A fire hydrant in the zone sits at elevation 1,032.0 feet, and the tank's water surface sits at elevation 1,186.0 feet. How many feet of static head are on that hydrant?
  2. Using 0.433 psi per foot of head, what is the static pressure at that hydrant, in psi, rounded to the nearest tenth?
  3. The customer who called sits at elevation 1,110.0 feet, and the tank's water surface is still at 1,186.0 feet. What is the static pressure at that house, in psi, rounded to the nearest tenth?
  4. The zone must hold at least 40 psi. Using 2.31 feet per psi and the tank water surface of 1,186.0 feet, what is the highest ground elevation, in feet, that still receives exactly 40 psi?
  5. The operator opens the 12-inch main, which is 1.0 foot across, and the water moves at 4.0 feet per second. Using area equals 0.785 times diameter squared, and flow equals area times velocity, what is the flow in cubic feet per second?
  6. Convert that flow of 3.14 cubic feet per second to gallons per minute, using 448.8 gallons per minute per cubic foot per second. Round to the nearest whole gpm.
  7. Convert that flow, carried to one decimal place as 1,409.2 gallons per minute, to million gallons per day, using 694 gallons per minute per MGD. Round to the nearest hundredth.
  8. The crew will flush a 2,400-foot run of that same 12-inch main, which is 1.0 foot across. Using 0.785 times diameter squared times length for the volume in cubic feet, and 7.48 gallons per cubic foot, how many gallons does that run of pipe hold?
  9. The crew flushes those 14,092 gallons at 1,409.2 gallons per minute. How many minutes does one full pipe volume take?

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key
  1. Answer: 154.0 ft. Static head equals the water surface elevation minus the point elevation. 1,186.0 ft minus 1,032.0 ft equals 154.0 ft of head.
  2. Answer: 66.7 psi. Multiply head by 0.433 psi per foot. 154.0 ft times 0.433 equals 66.682 psi, which rounds to 66.7 psi.
  3. Answer: 32.9 psi. The head on the house is 1,186.0 ft minus 1,110.0 ft, or 76.0 ft. Then 76.0 ft times 0.433 equals 32.908 psi, which rounds to 32.9 psi. This sits below the 40 psi the zone must hold, so the complaint is real and not a meter problem.
  4. Answer: 1,093.6 ft. A pressure of 40 psi needs 40 times 2.31, or 92.4 ft of head. Subtract that from the water surface: 1,186.0 ft minus 92.4 ft equals elevation 1,093.6 ft. Any ground above that elevation receives less than 40 psi.
  5. Answer: 3.14 cfs. Area equals 0.785 times 1.0 ft times 1.0 ft, or 0.785 square feet. Flow equals 0.785 square feet times 4.0 feet per second, which equals 3.14 cubic feet per second.
  6. Answer: 1,409 gpm. Multiply 3.14 cubic feet per second by 448.8 gallons per minute per cubic foot per second. That equals 1,409.232, which rounds to 1,409 gpm.
  7. Answer: 2.03 MGD. Divide 1,409.2 gallons per minute by 694 gallons per minute per MGD. That equals 2.0305, which rounds to 2.03 MGD.
  8. Answer: 14,092 gallons. Volume equals 0.785 times 1.0 ft times 1.0 ft times 2,400 ft, or 1,884 cubic feet. Multiply by 7.48 gallons per cubic foot: 1,884 times 7.48 equals 14,092.32, which rounds to 14,092 gallons.
  9. Answer: 10.0 minutes. Divide the volume by the flow: 14,092 gallons divided by 1,409.2 gallons per minute equals 10.0 minutes. The same answer comes from dividing the 2,400-foot run by the 4.0 feet per second velocity, which gives 600 seconds.

Differentiation

Support

Give students a labeled side view of the zone showing the tank water surface at elevation 1,186 feet, the hydrant, and the house, with the vertical distances marked, so the subtraction in problems 1 and 3 is visible before it is calculated. Print the two pressure constants at the top of the page and allow a calculator throughout.

Extension

Ask students to find the static pressure at a proposed new house at elevation 1,150.0 feet in the same fictional zone, then to state how many feet the tank's water surface would have to rise to bring that house to 40 psi, and to comment on whether raising a tank is a realistic fix.

ELL supports

Pre-teach head, static pressure, elevation, and flushing using the same side-view drawing, with an arrow showing that a lower address receives more pressure. Allow answers to be written as a number and a unit with the work shown in symbols rather than in sentences.

Extensions

Students can repeat problems 5 through 7 for an 8-inch main, which is 0.667 feet across, at the same 4.0 feet per second, and then explain to a partner why the flow drops by more than a third even though the diameter drops by only a third. A local utility's distribution supervisor could be asked which of these calculations appears most often in a working shift.